Limits When Determining Area between two Graphs

[Struggling]

Hi all having a little problem with finding the limits on the area between 2 graphs.

i can find the easy one such as:

Find the area between y=x^2 and y = 2x
which is:
x^2 = 2x
x^2 - 2x = 0
x(x-2) = 0

x = 0 & 2

but when i have a question like:
Find the area between y=2-x^2 & y =x

i can't work it out i got to x(1+x)= 2

but I am sooo lost
any help appreciated

 

[whozum]

Nvermind, I understand what your saying. To find the points of intersection between those two graphs, set them equal to each other.

2-x^2 = x

x^2 + x = 2

An obvious one is x=1.

Try quadratic formula.

 

[Struggling]

sorry whozum i don't think i explained the question well, i need to work out the points of intersection i have no problems working out the area.

yeh I've already got 1. so using the quadratic formula i should be able to find the points out?

 

[Struggling]

so the intersecting points are -2 & 1?

 

[whozum]

There you go. Graph it to make sure.

 

[steven187]

hello there

well first of all you need to find where both functions actually intersect this is done by making 2-x^2=x then using the quadratic formulae to find where they intersect, and so you will find that they will intersect at 1 and at -2 now if you want to find the area between these functions its best that you graph it and then split up the area which should correspond to the addition to a couple of integrals
\int_0^1 2-x-x^2 dx+\int_{-\sqrt{2}}^0 2-x^2+x dx-\int_{-2}^{-\sqrt 2} x -2+x^2 dx
by integrating you will be able to find the area between those two functions?
by the way y=2-x^2 has roots at +/-sqrt{2}
the area is 2.5 units hopefully without any small errors

 

[Struggling]

thanks guys!

 

[HallsofIvy]

hello there

well first of all you need to find where both functions actually intersect this is done by making 2-x^2=x then using the quadratic formulae to find where they intersect, and so you will find that they will intersect at 1 and at -2 now if you want to find the area between these functions its best that you graph it and then split up the area which should correspond to the addition to a couple of integrals
\int_0^1 2-x-x^2 dx+\int_{-\sqrt{2}}^0 2-x^2+x dx-\int_{-2}^{-\sqrt 2} x -2+x^2 dx
by integrating you will be able to find the area between those two functions?
by the way y=2-x^2 has roots at +/-sqrt{2}
the area is 2.5 units hopefully without any small errors

Why in the world should one do such a thing? For all x between -2 and 1, 2- x² is larger than x so 2-x^{2- x is positive and is the "height" of a thin rectangle between the two. The area is
\int_{-2}^1 2- x- x^2 dx= \frac{9}{2}= 4.5.}

 

[Struggling]

yeh that's tha answer i got 9/2